You want to prove,
$\displaystyle n^5-n$ is divisible by 5 (which is true this is Fermat's Little Theorem).

You can factor,
$\displaystyle n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)$

A number has 5 different forms,
$\displaystyle 5k,5k+1,5k+2,5k+3,5k+4$
If $\displaystyle n=5k$ that is clearly true because it has a factor $\displaystyle n=5k$. If $\displaystyle n=5k+1$ that is true because it has factor $\displaystyle n-1=5k$. If $\displaystyle n=5k+4$ that is true because it has factor $\displaystyle n+1=5k+4=5(k+1)$. Now it remains to show that $\displaystyle n=5k+2,5k+3=5k\pm 2$ when squared is $\displaystyle 5k+4$ add 1 in $\displaystyle n^2+1$ and you get the form of $\displaystyle 5k$